ruplun wrote:Kevin flips a coin 4 times .What is the probability that he gets heads on at least one of the 4 flips?
We can solve this, we'll using the complement (complements are often useful for questions involving "at least").
So, P(at least 1 head) = 1 -
P(no heads)
P(no heads) = P(all tails)
Aside: we'll use T1 to denote tails on first flip, T2 for tails on 2nd flip and so on.
P(all tails) = P(T1 AND T2 AND T3 AND T4)
P(all tails) = P(T1) x P(T2) x P(T3) x P(T4)
P(all tails) = 1/2 x 1/2 x 1/2 x 1/2
P(all tails) = 1/16
So, P(at least 1 head) = 1 -
P(no heads)
P(at least 1 head) = 1 -
1/16
P(at least 1 head) = 15/16
Cheers,
Brent
